Prove a formula for calculating the index of a specific subgroup of the special linear group SL_2(ℤ) | Step-by-Step Solution

Understanding the Index Formula for Γ₀(N)

What We're Solving: We need to prove that the index of the congruence subgroup Γ₀(N) in SL₂(ℤ) equals N · ∏(p|N)(1 + 1/p), where the product is over all prime divisors p of N.

The Approach: This is a beautiful application of counting techniques in group theory! We'll use the fundamental principle that |G| = |H| · [G:H] for any subgroup H of a finite group G. Since SL₂(ℤ) is infinite, we'll work with finite quotients and use properties of modular arithmetic. The key insight is to count matrices in SL₂(ℤ/Nℤ) that satisfy the Γ₀(N) condition.

Step-by-Step Solution:

Step 1: Set up the problem

• Recall that Γ₀(N) = {(a b; c d) ∈ SL₂(ℤ) : c ≡ 0 (mod N)}
• We want [SL₂(ℤ):Γ₀(N)] = |SL₂(ℤ/Nℤ)| / |Γ₀(N) mod N|

Step 2: Count total matrices in SL₂(ℤ/Nℓ) |SL₂(ℤ/Nℤ)| = N³ · ∏(p|N)(1 - 1/p²) comes from counting matrices (a b; c d) with ad - bc ≡ 1 (mod N).

Step 3: Count matrices in Γ₀(N) modulo N For matrices in Γ₀(N), we have c ≡ 0 (mod N), so our matrices look like (a b; 0 d) with ad ≡ 1 (mod N).

• Choose a: φ(N) choices (units modulo N)
• Choose d: determined by a since ad ≡ 1 (mod N)
• Choose b: N choices (any value modulo N)

So |Γ₀(N) mod N| = N · φ(N) = N² · ∏(p|N)(1 - 1/p)

Step 4: Apply the Chinese Remainder Theorem Since the formula is multiplicative, we can work prime by prime and use:

• For prime powers: [SL₂(ℤ):Γ₀(p^k)] = p^k(1 + 1/p)
• For general N: use multiplicativity of the index

Step 5: Calculate the index [SL₂(ℤ):Γ₀(N)] = |SL₂(ℤ/Nℤ)| / |Γ₀(N) mod N| = [N³ · ∏(p|N)(1 - 1/p²)] / [N² · ∏(p|N)(1 - 1/p)] = N · ∏(p|N)[(1 - 1/p²)/(1 - 1/p)] = N · ∏(p|N)[(p² - 1)/(p²)] · [p/(p - 1)] = N · ∏(p|N)(1 + 1/p)

The Answer: The formula [SL₂(ℤ):Γ₀(N)] = N · ∏(p|N)(1 + 1/p) is proven! The key steps were counting matrices with the congruence condition c ≡ 0 (mod N) and using the multiplicative structure of modular arithmetic.

Memory Tip: Remember this as "N times a correction factor for each prime." The factor (1 + 1/p) accounts for how the congruence condition c ≡ 0 (mod p) restricts our choices. It's bigger than 1 because we're measuring how much smaller Γ₀(N) is compared to the full group!

The beauty here is how number theory (multiplicative functions) connects with group theory (subgroup indices). Keep practicing with small examples like N = 6 to build intuition! 🌟